In a quadrilateral $ABCD$, we have $\angle ACB = \angle ADB = 90$ and $CD < BC$. Denote $E$ as the intersection of $AC$ and $BD$, and let the perpendicular bisector of $BD$ hit $BC$ at $F$. The circle with center $F$ which passes through $B$ hits $AB$ at $P (\neq B)$ and $AC$ at $Q$. Let $M$ be the midpoint of $EP$. Prove that the circumcircle of $EPQ$ is tangent to $AB$ if and only if $B, M, Q$ are colinear.
Problem
Source: 2017 KMO Problem 6
Tags: geometry, circumcircle
12.11.2017 13:28
Darn!!!!! It’s impossible for me to get full score. By the way, when we get that $E$ is orthocenter of some triangle, it can be very easy problem!
12.11.2017 13:46
Huh. Here's a sketch of my solution.
12.11.2017 14:18
Consider point $X=\overline{AD} \cap \overline{BC}$; then $E$ is the orthocenter of $\triangle XAB$. Also $P=\overline{XE} \cap \overline{AB}$ and $F$ is the midpoint of $\overline{XB}$. Pick point $R$ on line $\overline{AC}$ extended beyond $C$ with $\angle XRB=90^{\circ}$. Claim. $\odot(EPQ)$ touches $\overline{AB}$ if and only if $\overline{XR} \parallel \overline{AB}$ if and only if $B, M, Q$ are collinear. (Proof) For (1), note that $\odot(EPQ)$ touch $\overline{AB}$ iff $\angle EQP=90^{\circ}$ iff $\angle PXR=\angle PQR=90^{\circ}$ iff $\overline{XR} \parallel \overline{AB}$. For (2), note that $CQ \cdot CR=CE \cdot CA$ so $(A, C, Q, R)=-1$. Project it from $B$ onto $\overline{XE}$ and we get that $\angle RBA=90^{\circ}$ as desired. $\blacksquare$
29.07.2018 11:12
Suppose that $(QEP)$ is tangent to $AB$. $\angle QEP = \angle QPA = \angle QBD = \frac {1} {2} \angle QFD = 90 - \angle CDQ = \angle CQD (1)$ $\angle PBD = \frac {1} {2} \angle PFD = 90 - \angle FPD = \angle FAP$ $\longrightarrow BEPA$ is cyclic $\longrightarrow \angle EPD = 90$ $\longrightarrow \angle EQP = 90 , (1) \longrightarrow QD$ is median of $EP.$ And proof of the other side is similar to above.
07.03.2019 13:39
Let $AD$ and $BC$ intersect at $X$, then $E$ is the orthocentre of $AXB$. Furthermore, $F$ is the midpoint of $BX$ as $X$ is the midpoint of $BD$ and $\angle BXF=\angle BDX=90$. So, the diameter of the circle through $F$ is $BX$. Therefore $\angle BPX=90$ too, so $P,E,X$ are collinear ($PX$ is altitude). If $PQE$ is tangent to $AB$: $\angle PQE=90 \implies M$ is the circumcentre of $PQE \implies \angle MQP=\angle MPQ$, so as $BX \parallel PQ$, $\angle MPQ = \angle PXB = \angle BQP$ therefore $Q,M,B$ collinear. Now, if $Q,M,B$ collinear: let $\angle PQM=a$, $\angle MPQ=b$. Then $\angle QEM = \angle XEC=90-\angle PXB=90-a$. We also have $\angle MQE = 180-a-b-(90-a)=90-b$. Then $$\frac{EM}{\sin(90-b)}=\frac{QM}{\sin(90-a)}$$and $$\frac{PM}{\sin a}=\frac{QM}{\sin b}$$Combining and making use of $EM=PM$: $$\frac{\sin a}{\sin(90-a)}=\frac{\sin b}{\sin(90-b)}$$But as $0\le a,b \le 90$ the only such solution is $a=b$, therefore $QM=MP$ and $\angle PQE=90$ as desired.
02.07.2021 19:31
Sorry for 2 year bump, but why do trig bash the 2nd part@above?its pretty easy ,see that MQ=MP=ME,which will give PQE = 90 degrees and done!!
30.06.2022 18:22
Let $G=AD\cap BC$. Still have to prove (QGE) is tangent to GC :sad. I will edit later. Claim: $G$ lies on $(BPQ)$. Proof. Obviously $B$ lies on $(BPQ)$. Let $O$ be the midpoint of $AB$. So $AD\parallel OF$ because both are perpendicular to $BD$. Since $O$ be the midpoint of $AB$ therefore $F$ is the midpoint of $GB$. $\blacksquare$ So we can say $E$ is the orthocenter of $\triangle ABG$. Therefore $PE\perp AB$. Claim (if): $(PQE)$ is tangent to $AB$ if $B,M,Q$ are collinear. Proof. Notice that $GQ$ is tangent to $(CBQ)$ because $\angle CBQ=\angle CQG$. Now \[ \angle EPQ=\angle GPQ=\angle CGE \]which means $PQ\parallel CG$ and hence $\angle PQE=90^\circ.$ $\blacksquare$ Claim (only if): $B,M,Q$ are collinear if $(PQE)$ is tangent to $AB$. Proof. So the $\angle PQE=90^\circ$ hence $PQGB$ is isosceles trapezoid and so $\angle PQM=\angle MPQ=\angle GPQ=\angle PQB$. $\blacksquare$